To determine which point has a distance of 10 units from the point \((-1, 2)\), we can use the distance formula between two points \((x_1, y_1)\) and \((x_2, y_2)\):
\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \]
In this case, \((x_1, y_1) = (-1, 2)\). We want to find a point \((x_2, y_2)\) such that the distance \(d = 10\).
We can set up the equation:
\[ \sqrt{(x_2 - (-1))^2 + (y_2 - 2)^2} = 10 \]
Squaring both sides:
\[ (x_2 + 1)^2 + (y_2 - 2)^2 = 100 \]
Now, we check each of the given points to see which satisfies this equation.
-
Point \((-11, 2)\): \[ (-11 + 1)^2 + (2 - 2)^2 = (-10)^2 + 0^2 = 100 + 0 = 100 \quad \text{(satisfies)} \]
-
Point \((-1, -12)\): \[ (-1 + 1)^2 + (-12 - 2)^2 = 0^2 + (-14)^2 = 0 + 196 = 196 \quad \text{(does not satisfy)} \]
-
Point \((-1, 10)\): \[ (-1 + 1)^2 + (10 - 2)^2 = 0^2 + 8^2 = 0 + 64 = 64 \quad \text{(does not satisfy)} \]
-
Point \((10, 2)\): \[ (10 + 1)^2 + (2 - 2)^2 = 11^2 + 0^2 = 121 + 0 = 121 \quad \text{(does not satisfy)} \]
The only point that has a distance of 10 from the point \((-1, 2)\) is:
\[ \boxed{(-11, 2)} \]
1 answer